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177 geometry/dissections/largest.circle.p




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This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

177 geometry/dissections/largest.circle.p


What is the largest circle that can be assembled from two semicircles cut from
a rectangle with edges a and b?

geometry/dissections/largest.circle.s

There are two methods:

Method M1:
The diameters of the semicircles have to be on the longer sides,
starting at an endpoint of the rectangle. The two semicircles touch
each other in the middle M of the rectangle.

		    a
       D._______________________.C
	|			|
	|			|
      b	|	    . M		|
	|			|
	|			|
	|_______.___.___________|
	A       R   X		B

R should be the center of the semicircle, and because of RA = RM,
it holds that:
r^2 = (a/2 - r)^2 + (b/2)^2

Solving for r gives:

r = min[b,(a^2+b^2)/(4a)], where a >= b.

Method M2:
We'll cut on the line y = c x, where c will turn out to be slightly
less than d, the slope of the diagonal. We describe the semicircle
lying above the line y = c x, having this line as the straight part of
the semi-circle. The center P of the semicircle will be taken on the
line y = d - x, and will be tangent to the left and top of the
rectangle. Clearly the lower down P is on this line, the better. The
naive solution is not optimal because the upper place where the
semicircle meets the diagonal is interior to the rectangle. So we try
to determine c in such a way that this latter point actually lies
slightly down from the top, on the right side of the rectangle. This
involves solving the quartic:
4r^4 - (4a+16b)r^3 + (16b^2+a^2+8ab)r^2 - (6b^3+4ab^2+2ba^2)r + b^4+(ab)^2 = 0,
where r < b, the details of which will be left to the reader.

The other semicircle is the reflection of the first through the origin.

After a few calculations, we find that the value of r given
by M2 is greater than the one given by M1 only if 1 < a/b < 2.472434.

 

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